The standard electromotive force emf for an electrochemical reaction is

The standard electromotive force emf for an

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The standard electromotive force (emf) for an electrochemical reaction is calculated simply as the difference between the standard reduction potentials between the species being reduced and the species being oxidized: E ° = 0 red E (Species undergoing reduction) 0 red E (Species undergoing oxidation); or E ° = 0 red E (Cathode half-cell reaction) 0 red E (Anode half-cell reaction). Example : Using the table of standard reduction potentials, what is the standard electrochemical potential ( E ° , emf in V), standard free energy change ( G ° in kJ) and K c at 298 K for the reaction Zn(s) + Cu 2+ (aq) ⎯→ Zn 2+ (aq) + Cu(s)? In this reaction, electrons are transferred from Zn(s) (it is oxidized) to Cu 2+ (aq) (it is reduced). The standard electrochemical potential is the difference between the standard reduction potential for the couple that receives the electrons and the standard reduction potential for the couple which gives away the electrons: E ° = 0 red E (Cu 2+ /Cu) 0 red E (Zn/Zn 2+ ) = +0.34 V ( 0.76 V) = +1.10 V (spontaneous!) For the standard free energy change, the number of electrons transferred is 2 moles (per mole of Zn). Therefore, G ° = nF E ° = (2 moles)(96,500 coulomb/mole)(+1.10 V) = 212.3 kJ K c = exp( −∆ G ° / RT ) = exp(85.65) = 1.57 × 10 37 (a very large number!!) = 2 2 [Zn ] [Cu ] + + . Therefore, when equilibrium is achieved, we expect to observe essentially no Cu 2+ (aq). So, we do not use an equilibrium symbol, but just an arrow because the concentration of Cu 2+ (aq) would be immeasurable. Example : What is the standard electrochemical potential (emf), the standard free energy change, and K c for the reaction Ag(s) + Fe 2+ (aq) ⎯→ Ag + (aq) + Fe(s)? In this reaction, electrons are transferred from Ag(s) (it is oxidized) to Fe 2+ (aq) (it is reduced). The reaction must first be balanced: 2 Ag(s) + Fe 2+ (aq) ⎯→ 2 Ag + (aq) + Fe(s) E ° is the potential energy change PER coulomb of charge transferred. Therefore, we can simply take the difference in 0 red E values...
E ° = 0 red E (Fe 2+ /Fe) 0 red E (Ag/Ag + ) = 0.44 V (+0.80 V) = 1.24 V (not spontaneous!) For the standard free energy change, the number of electrons transferred is 2 moles (per mole of Fe). Therefore, G ° = nF E ° = (2 moles)(96,500 coulomb/mole)( 1.24 V) = +239.3 kJ K c = exp( −∆ G ° / RT ) = exp( 96.59) = 1.13 × 10 42 (a very small number!!) = 2 2 [Ag ] [Fe ] + + . Reduction-Potential Diagrams : Numerous chemical elements exhibit different oxidation states in their chemical species. These elements include many of the transition metals (V, Cr, Mn, Fe, Co, Ni, Cu) and some of the main group elements (N, O). Standard reduction potentials between different oxidation states of these elements can be succinctly summarized in a reduction-potential diagram . An example of one is shown here for Cu, which shows common oxidation states of Cu 0 , Cu + , and Cu 2+ in aqueous media: In this diagram, we can write down the following three reduction half-reactions: (1) Cu 2+ (aq) + e ⎯→ Cu + (aq) 0 red E = +0.158 V; 0 red G = 15.25 kJ (2) Cu + (aq) + e ⎯→ Cu(aq) 0 red E = +0.522 V; 0 red G = 50.37 kJ (3) Cu 2+ (aq) + 2e ⎯→ Cu(aq) 0 red E = +0.340 V; 0 red G = 65.62 kJ Note that adding half-reactions (1) and (2) together give half-reaction (3), but if we add the corresponding 0 red E values, we get +0.680 V.

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• Fall '07
• Miller

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